A rod of mass M and length l is suspended by | vedclass

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(D) The forces acting on the rod are:$1$. Gravitational force $Mg$ acting downwards.$2$. Magnetic force $F_m = BIl$ acting upwards (by Fleming's Left-

Vedclass · Accepted Answer

$\frac{Mg-BIl}{2}$

Vedclass · Answer

(D) The forces acting on the rod are:$1$. Gravitational force $Mg$ acting downwards.$2$. Magnetic force $F_m = BIl$ acting upwards (by Fleming's Left-Hand Rule,as current flows to the right and field is into the page).$3$. Tension $T$ in each of the two wires acting upwards.For the rod to be in equilibrium,the net force must be zero:$2T + F_m = Mg$$2T + BIl = Mg$$2T = Mg - BIl$$T = \frac{Mg - BIl}{2}$

$A$ rod of mass $M$ and length $l$ is suspended by two wires $P$ and $Q$. $A$ uniform magnetic field $B$ is directed into the page. When current $I$ flows through the rod,the tension in each supporting wire is:

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